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From: Oscar Benjamin <oscar.j.benjamin@gmail.com>
Date: Tue, 4 Mar 2014 22:08:44 +0000
Subject: Re: Working with the set of real numbers
To: Marko Rauhamaa <marko@pacujo.net>
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Xref: csiph.com comp.lang.python:67739

On 4 March 2014 21:05, Marko Rauhamaa <marko@pacujo.net> wrote:
> Oscar Benjamin <oscar.j.benjamin@gmail.com>:
>
>> To me the obvious method is Newton iteration which takes O(sqrt(N))
>> iterations to obtain N digits of precision. This brings the above
>> complexity below quadratic:
>>
>> #!/usr/bin/env python
>>
>> from decimal import Decimal as D, localcontext
>>
>> def sqrt(y, prec=3D1000):
>>     '''Solve x**2 =3D y'''
>>     assert y > 0
>>     eps =3D D(10) ** -(prec + 5)
>>     x =3D D(y)
>>     with localcontext() as ctx:
>>         ctx.prec =3D prec + 10
>>         while x ** 2 - y > x * eps:
>>             x =3D (x + y/x) / 2
>>     return x
>>
>> print(sqrt(2))
>
> At a quick glance, I believe x ** 2 is O(N=B2) and so the total complexit=
y
> should be O(N ** 2.5).

x**2 is just a multiplication which can be done in better than O(N**2):
http://en.wikipedia.org/wiki/Multiplication_algorithm#Fast_multiplication_a=
lgorithms_for_large_inputs


Oscar